Integrand size = 21, antiderivative size = 430 \[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx=-\frac {a b x}{c e}-\frac {b^2 x \arctan (c x)}{c e}-\frac {i d (a+b \arctan (c x))^2}{c e^2}+\frac {(a+b \arctan (c x))^2}{2 c^2 e}-\frac {d x (a+b \arctan (c x))^2}{e^2}+\frac {x^2 (a+b \arctan (c x))^2}{2 e}-\frac {d^2 (a+b \arctan (c x))^2 \log \left (\frac {2}{1-i c x}\right )}{e^3}-\frac {2 b d (a+b \arctan (c x)) \log \left (\frac {2}{1+i c x}\right )}{c e^2}+\frac {d^2 (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^3}+\frac {b^2 \log \left (1+c^2 x^2\right )}{2 c^2 e}+\frac {i b d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right )}{e^3}-\frac {i b^2 d \operatorname {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{c e^2}-\frac {i b d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^3}-\frac {b^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e^3}+\frac {b^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^3} \]
-a*b*x/c/e-b^2*x*arctan(c*x)/c/e-I*d*(a+b*arctan(c*x))^2/c/e^2+1/2*(a+b*ar ctan(c*x))^2/c^2/e-d*x*(a+b*arctan(c*x))^2/e^2+1/2*x^2*(a+b*arctan(c*x))^2 /e-d^2*(a+b*arctan(c*x))^2*ln(2/(1-I*c*x))/e^3-2*b*d*(a+b*arctan(c*x))*ln( 2/(1+I*c*x))/c/e^2+d^2*(a+b*arctan(c*x))^2*ln(2*c*(e*x+d)/(c*d+I*e)/(1-I*c *x))/e^3+1/2*b^2*ln(c^2*x^2+1)/c^2/e+I*b*d^2*(a+b*arctan(c*x))*polylog(2,1 -2/(1-I*c*x))/e^3-I*b^2*d*polylog(2,1-2/(1+I*c*x))/c/e^2-I*b*d^2*(a+b*arct an(c*x))*polylog(2,1-2*c*(e*x+d)/(c*d+I*e)/(1-I*c*x))/e^3-1/2*b^2*d^2*poly log(3,1-2/(1-I*c*x))/e^3+1/2*b^2*d^2*polylog(3,1-2*c*(e*x+d)/(c*d+I*e)/(1- I*c*x))/e^3
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx=\int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx \]
Time = 0.70 (sec) , antiderivative size = 430, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {5411, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx\) |
| \(\Big \downarrow \) 5411 |
| \(\displaystyle \int \left (\frac {d^2 (a+b \arctan (c x))^2}{e^2 (d+e x)}-\frac {d (a+b \arctan (c x))^2}{e^2}+\frac {x (a+b \arctan (c x))^2}{e}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(a+b \arctan (c x))^2}{2 c^2 e}+\frac {i b d^2 \operatorname {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) (a+b \arctan (c x))}{e^3}-\frac {i b d^2 (a+b \arctan (c x)) \operatorname {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^3}-\frac {d^2 \log \left (\frac {2}{1-i c x}\right ) (a+b \arctan (c x))^2}{e^3}+\frac {d^2 (a+b \arctan (c x))^2 \log \left (\frac {2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^3}-\frac {d x (a+b \arctan (c x))^2}{e^2}-\frac {i d (a+b \arctan (c x))^2}{c e^2}-\frac {2 b d \log \left (\frac {2}{1+i c x}\right ) (a+b \arctan (c x))}{c e^2}+\frac {x^2 (a+b \arctan (c x))^2}{2 e}-\frac {a b x}{c e}-\frac {b^2 x \arctan (c x)}{c e}+\frac {b^2 \log \left (c^2 x^2+1\right )}{2 c^2 e}-\frac {b^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 e^3}+\frac {b^2 d^2 \operatorname {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^3}-\frac {i b^2 d \operatorname {PolyLog}\left (2,1-\frac {2}{i c x+1}\right )}{c e^2}\) |
-((a*b*x)/(c*e)) - (b^2*x*ArcTan[c*x])/(c*e) - (I*d*(a + b*ArcTan[c*x])^2) /(c*e^2) + (a + b*ArcTan[c*x])^2/(2*c^2*e) - (d*x*(a + b*ArcTan[c*x])^2)/e ^2 + (x^2*(a + b*ArcTan[c*x])^2)/(2*e) - (d^2*(a + b*ArcTan[c*x])^2*Log[2/ (1 - I*c*x)])/e^3 - (2*b*d*(a + b*ArcTan[c*x])*Log[2/(1 + I*c*x)])/(c*e^2) + (d^2*(a + b*ArcTan[c*x])^2*Log[(2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x) )])/e^3 + (b^2*Log[1 + c^2*x^2])/(2*c^2*e) + (I*b*d^2*(a + b*ArcTan[c*x])* PolyLog[2, 1 - 2/(1 - I*c*x)])/e^3 - (I*b^2*d*PolyLog[2, 1 - 2/(1 + I*c*x) ])/(c*e^2) - (I*b*d^2*(a + b*ArcTan[c*x])*PolyLog[2, 1 - (2*c*(d + e*x))/( (c*d + I*e)*(1 - I*c*x))])/e^3 - (b^2*d^2*PolyLog[3, 1 - 2/(1 - I*c*x)])/( 2*e^3) + (b^2*d^2*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + I*e)*(1 - I*c*x)) ])/(2*e^3)
3.2.42.3.1 Defintions of rubi rules used
Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f* x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[p, 0] & & IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 26.05 (sec) , antiderivative size = 1710, normalized size of antiderivative = 3.98
| method | result | size |
| parts | \(\text {Expression too large to display}\) | \(1710\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1718\) |
| default | \(\text {Expression too large to display}\) | \(1718\) |
1/2*a^2/e*x^2-a^2/e^2*x*d+a^2/e^3*d^2*ln(e*x+d)+b^2/c^3*(1/2*arctan(c*x)^2 /e*c^3*x^2-arctan(c*x)^2/e^2*c^3*d*x+c^3*arctan(c*x)^2*d^2/e^3*ln(c*e*x+c* d)-2*c*(-1/4/e*arctan(c*x)^2-1/2*I/e^3*c^2*d^2*arctan(c*x)*polylog(2,-(1+I *c*x)^2/(c^2*x^2+1))-1/4*I/e^3*csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1 +I*c*x)^2/(c^2*x^2+1)+I*e+c*d)/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(-I*e*( 1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d))*csgn(I/((1+I* c*x)^2/(c^2*x^2+1)+1))*Pi*c^2*d^2*arctan(c*x)^2-1/2*I/e^2*c*d*arctan(c*x)^ 2+1/4*I/e^3*c^2*d^2*Pi*csgn(I/((1+I*c*x)^2/(c^2*x^2+1)+1))*csgn(I*(-I*e*(1 +I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I*e+c*d)/((1+I*c*x)^2/(c ^2*x^2+1)+1))^2*arctan(c*x)^2-I/e^2*c*d*dilog(1+I*(1+I*c*x)/(c^2*x^2+1)^(1 /2))+1/4*I/e^3*c^2*d^2*Pi*csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c* x)^2/(c^2*x^2+1)+I*e+c*d))*csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c *x)^2/(c^2*x^2+1)+I*e+c*d)/((1+I*c*x)^2/(c^2*x^2+1)+1))^2*arctan(c*x)^2+1/ 2/e*ln((1+I*c*x)^2/(c^2*x^2+1)+1)+1/2*arctan(c*x)*(c*x-I)/e-1/4*I/e^3*Pi*c ^2*d^2*csgn(I*(-I*e*(1+I*c*x)^2/(c^2*x^2+1)+c*d*(1+I*c*x)^2/(c^2*x^2+1)+I* e+c*d)/((1+I*c*x)^2/(c^2*x^2+1)+1))^3*arctan(c*x)^2+1/e^2*c*d*arctan(c*x)* ln(1+I*(1+I*c*x)/(c^2*x^2+1)^(1/2))+1/e^2*c*d*arctan(c*x)*ln(1-I*(1+I*c*x) /(c^2*x^2+1)^(1/2))+1/2*I*d^3*c^3/e^3/(c*d-I*e)*arctan(c*x)*polylog(2,(I*e -c*d)/(c*d+I*e)*(1+I*c*x)^2/(c^2*x^2+1))+1/4/e^3*c^2*d^2*polylog(3,-(1+I*c *x)^2/(c^2*x^2+1))+1/2*d^2*c^2/e^3*arctan(c*x)^2*ln(-I*e*(1+I*c*x)^2/(c...
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{e x + d} \,d x } \]
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx=\int \frac {x^{2} \left (a + b \operatorname {atan}{\left (c x \right )}\right )^{2}}{d + e x}\, dx \]
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{e x + d} \,d x } \]
1/2*a^2*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + 1/32*(4*(b^2*e*x^ 2 - 2*b^2*d*x)*arctan(c*x)^2 + 32*e^2*integrate(1/16*(12*(b^2*c^2*e^2*x^4 + b^2*e^2*x^2)*arctan(c*x)^2 + (b^2*c^2*e^2*x^4 + b^2*e^2*x^2)*log(c^2*x^2 + 1)^2 + 4*(8*a*b*c^2*e^2*x^4 - b^2*c*e^2*x^3 + 2*b^2*c*d^2*x + (b^2*c*d* e + 8*a*b*e^2)*x^2)*arctan(c*x) + 2*(b^2*c^2*e^2*x^4 - b^2*c^2*d*e*x^3 - 2 *b^2*c^2*d^2*x^2)*log(c^2*x^2 + 1))/(c^2*e^3*x^3 + c^2*d*e^2*x^2 + e^3*x + d*e^2), x) - (b^2*e*x^2 - 2*b^2*d*x)*log(c^2*x^2 + 1)^2)/e^2
\[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx=\int { \frac {{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{e x + d} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \arctan (c x))^2}{d+e x} \, dx=\int \frac {x^2\,{\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^2}{d+e\,x} \,d x \]